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Difference between revisions of "1997 AHSME Problems/Problem 9"

(Created page with "==Problem 9== In the figure, <math>ABCD</math> is a <math>2 \times 2</math> square, <math>E</math> is the midpoint of <math>\overline{AD}</math>, and <math>F</math> is on <math>...")
 
 
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<math> \textbf{(A)}\ 2\qquad\textbf{(B)}\ 3-\frac{\sqrt{3}}{2}\qquad\textbf{(C)}\ \frac{11}{5}\qquad\textbf{(D)}\ \sqrt{5}\qquad\textbf{(E)}\ \frac{9}{4} </math>
 
<math> \textbf{(A)}\ 2\qquad\textbf{(B)}\ 3-\frac{\sqrt{3}}{2}\qquad\textbf{(C)}\ \frac{11}{5}\qquad\textbf{(D)}\ \sqrt{5}\qquad\textbf{(E)}\ \frac{9}{4} </math>
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==Solution 1==
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Since <math>\angle EBA = \angle FCB</math> and <math>\angle FBC = \angle AEB</math>, we have <math>\triangle ABE \sim \triangle FCB</math>.
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<math>\frac{AB}{FC} = \frac{BE}{CB} = \frac{EA}{BF}</math>
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<math>\frac{2}{FC} = \frac{\sqrt{5}}{2} = \frac{1}{BF}</math>
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From those two equations, we find that <math>CF = \frac{4}{\sqrt{5}}</math> and <math>BF = \frac{2}{\sqrt{5}}</math>
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Now that we have <math>BF</math> and <math>CF</math>, we can find the area of the bottom triangle <math>\triangle CFB</math>:  <math>\frac{1}{2}\cdot \frac{4}{\sqrt{5}} \cdot \frac{2}{\sqrt{5}} = \frac{4}{5}</math>
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The area of left triangle <math>\triangle BEA</math> is <math>\frac{1}{2}\cdot 2 \cdot 1 = 1</math>
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The area of the square is <math>4</math>.
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Thus, the area of the remaining quadrilateral is <math>4 - 1 - \frac{4}{5} = \frac{11}{5}</math>, and the answer is <math>\boxed{C}</math>
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==Solution 2==
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Place the square on a coordinate grid so that <math>B(0,0)</math> and <math>D(2,2)</math>.  Line <math>BE</math> is <math>y = 2x</math>.  Line <math>CF</math> goes through <math>(2,0)</math> and has slope <math>-\frac{1}{2}</math>, so it must be <math>y = 1 -\frac{x}{2}</math>
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The intersection of the two lines is <math>F</math>, and <math>F</math> thus has coordinates <math>(\frac{2}{5}, \frac{4}{5})</math>.  The altitude from <math>F</math> to <math>BC</math> thus has length <math>\frac{4}{5}</math>, so the area of the triangle <math>BCF</math> is <math>\frac{1}{2}\cdot 2\cdot \frac{4}{5} = \frac{4}{5}</math>.
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The other triangle has area <math>1</math>, and the whole square has area <math>4</math>.  As above, we find the area of the quadrilateral by subtracting the two triangles, and we get <math>\frac{11}{5}</math>, which is <math>\boxed{C}</math>.
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== See also ==
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{{AHSME box|year=1997|num-b=8|num-a=10}}
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{{MAA Notice}}

Latest revision as of 13:12, 5 July 2013

Problem 9

In the figure, $ABCD$ is a $2 \times 2$ square, $E$ is the midpoint of $\overline{AD}$, and $F$ is on $\overline{BE}$. If $\overline{CF}$ is perpendicular to $\overline{BE}$, then the area of quadrilateral $CDEF$ is

[asy] defaultpen(linewidth(.8pt)); dotfactor=4; pair A = (0,2); pair B = origin; pair C = (2,0); pair D = (2,2); pair E = midpoint(A--D); pair F = foot(C,B,E); dot(A);dot(B);dot(C);dot(D);dot(E);dot(F); label("$A$",A,N);label("$B$",B,S);label("$C$",C,S);label("$D$",D,N);label("$E$",E,N);label("$F$",F,NW); draw(A--B--C--D--cycle); draw(B--E); draw(C--F); draw(rightanglemark(B,F,C,4));[/asy]

$\textbf{(A)}\ 2\qquad\textbf{(B)}\ 3-\frac{\sqrt{3}}{2}\qquad\textbf{(C)}\ \frac{11}{5}\qquad\textbf{(D)}\ \sqrt{5}\qquad\textbf{(E)}\ \frac{9}{4}$

Solution 1

Since $\angle EBA = \angle FCB$ and $\angle FBC = \angle AEB$, we have $\triangle ABE \sim \triangle FCB$.

$\frac{AB}{FC} = \frac{BE}{CB} = \frac{EA}{BF}$

$\frac{2}{FC} = \frac{\sqrt{5}}{2} = \frac{1}{BF}$

From those two equations, we find that $CF = \frac{4}{\sqrt{5}}$ and $BF = \frac{2}{\sqrt{5}}$

Now that we have $BF$ and $CF$, we can find the area of the bottom triangle $\triangle CFB$: $\frac{1}{2}\cdot \frac{4}{\sqrt{5}} \cdot \frac{2}{\sqrt{5}} = \frac{4}{5}$

The area of left triangle $\triangle BEA$ is $\frac{1}{2}\cdot 2 \cdot 1 = 1$

The area of the square is $4$.

Thus, the area of the remaining quadrilateral is $4 - 1 - \frac{4}{5} = \frac{11}{5}$, and the answer is $\boxed{C}$

Solution 2

Place the square on a coordinate grid so that $B(0,0)$ and $D(2,2)$. Line $BE$ is $y = 2x$. Line $CF$ goes through $(2,0)$ and has slope $-\frac{1}{2}$, so it must be $y = 1 -\frac{x}{2}$

The intersection of the two lines is $F$, and $F$ thus has coordinates $(\frac{2}{5}, \frac{4}{5})$. The altitude from $F$ to $BC$ thus has length $\frac{4}{5}$, so the area of the triangle $BCF$ is $\frac{1}{2}\cdot 2\cdot \frac{4}{5} = \frac{4}{5}$.

The other triangle has area $1$, and the whole square has area $4$. As above, we find the area of the quadrilateral by subtracting the two triangles, and we get $\frac{11}{5}$, which is $\boxed{C}$.

See also

1997 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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